A package of mass m is placed onto a horizontal conveyor belt ...

Question

A package of mass m is placed onto a horizontal conveyor belt moving at speed v (Fig. 7–32). The coefficient of kinetic friction between package and belt is μₖ .

(a) How long does it take the package to stop sliding on the belt? [Hint: Use F = ma.]

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A crate weighing 50 kg is placed onto a flat bed of a carrier truck that is traveling horizontally at a constant speed of 15 m per as shown in the figure below the coefficient of kinetic friction UK between the crate and the flatbed is 0.2 determine the duration required for the crate to cease all motion after it's positioned on a on the moving flatbed. OK. So that's our end goal is we need to determine the duration required for the crate to cease all motion after it's positioned on the moving flatbed. So ultimately, what we're trying to solve for our final answers, we're trying to figure out what the time is. So what time is required for the crate to stop moving once it's placed on this moving flatbed? Awesome. So with that in mind, let's look at our figure here. So we have our driver wearing a green shirt in this carrier truck and it's moving to the right at 2.0 m per second. And then we have a gray box representing our crate that's in this flat bed of this carrier truck. So now that we have an idea visually what's going on in this pro and then of course, this black line that the carrier truck is laying on or resting on, standing on is denoting the ground or the road. So we are also given some multiple choice answers. And let's note that they're all in the same units of seconds. So let's read them off to see what our final answer might be. A is 1.02 B is 2.40 and C is 4.51 and D is 5.00. Awesome. So first off, let us create a free body diagram to help us better visualize the forces at work in this problem. So I've gone ahead and drawn a free body diagram for us to look at. So the black line at the very bottom here denotes the ground. And inside this one, we have two little boxes here. Our first box represents the carrier truck and then the second box denotes the crate. So our first force in the downwards direction is the mass of the crate multiplied by the gravitation like by the by gravity, which this is known as the gravitational force. So mass multiplied by gravity equals the gravitational force. And then since our truck is moving 2.0 m per second. In the right direction, we have our friction force in the opposing direction going to the left. And of course, in the upwards direction, we have our normal force which is denoted FN. And our friction force is f subscript Fr Awesome. So now let us note that the carrier truck is sliding underneath the crate to the right. Thus, there will be a force of the kinetic friction on the crate that will reach a speed of 2.0 m per second. So now we need to solve for the net force in the X direction by recalling and using the following equation that the sum of FX is equal to FFR. So if we recall and apply Newton's laws, we can write that and let's call this equation one that FFR is equal to M multiplied by A. So the friction force is equal to mass multiplied by acceleration. We can also write and let's call this equation two that FS so if FFR, so the friction force is equal to mu K, the kinetic friction coefficient multiplied by FN is equal to mu K multiplied by M multiplied by G. So we can now set equations one and two equal to each other. And we can write that M multiplied by A is equal to MU K multiplied by M multiplied by G. So note that M will cancel out and we will be then be able to rearrange this equation and isolate and sulfur a much easier once M cancels out. So when we do that, we will find that A, the acceleration is equal to mu K multiplied by G which is equal to 0.2 multiplied by 9.8 which is the numerical value for gravity and seen as our meters per second squared. So when you plug that into your calculator, you will get 1.96 m per second squared. So now we need to recall and use the uniform accelerated motion equations to help us solve for T, which is our final answer. So let's recall that VX equals V zero X plus A X multiplied by T. So we need to note that VX is our final velocity of the crate. And let's make a little note here off to the side. So VX equals 2.0 m per second because that's our final velocity of the crate and V zero X equals the initial velocity of the crate, which is 0 m per second because it starts from rest. So now at this point, we can plug in all of our known variables to solve for T. So we need to plug in over a variable. So VX is 2.0 m per second equals 0 m per second plus our acceleration, which is 1.96 m per second squared multiplied by T. So when we rearrange this equation to isolate and solve for T we will find that T is equal to 2.0 m per second, divided by 1.96 m per second squared. So when we plug that into our calculator and we round to two decimal places, we will get 1.02 seconds. And that is our final answer. Hooray. We did it. So looking at our multiple choice answers, the correct answer has to be the letter A 1.02 seconds. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Struggling with Physics?

A package of mass m is placed onto a horizontal conveyor belt moving at speed v (Fig. 7–32). The coefficient of kinetic friction between package and belt is μₖ .
<IMAGE>

(a) How long does it take the package to stop sliding on the belt? [Hint: Use F = ma.]

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Struggling with Physics?

A package of mass m is placed onto a horizontal conveyor belt moving at speed v (Fig. 7–32). The coefficient of kinetic friction between package and belt is μₖ .<IMAGE>(a) How long does it take the package to stop sliding on the belt? [Hint: Use F = ma.]

Watch next

Newton's First & Second Laws practice set

A package of mass m is placed onto a horizontal conveyor belt moving at speed v (Fig. 7–32). The coefficient of kinetic friction between package and belt is μₖ .
<IMAGE>

(a) How long does it take the package to stop sliding on the belt? [Hint: Use F = ma.]